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Math Help - Quick question

  1. #1
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    Quick question

    If \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)

    What is \left<x^2\right>?
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  2. #2
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    Quote Originally Posted by Captcha View Post
    If \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)

    What is \left<x^2\right>?
    so square both sides of your initial equation
     (x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2

    and evaluate further if you know what the x1,x2,....etc are
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  3. #3
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    Quote Originally Posted by purebladeknight View Post
    so square both sides of your initial equation
     (x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2

    and evaluate further if you know what the x1,x2,....etc are
    It's not \left<x\right>^2 I want to find. It's \left<x^2\right><br />
.

    Basically I'm trying to solve for x the equation:

    2.61 = \sqrt{\left<x^2\right>-\left<x\right>^2} to find the value of x, where 2.61 is the \text{standard deviation} of the data 8,7, 4, 10, 6,  x.

    I know that \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right).

     \Rightarrow \left<x^2\right> = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2.

    So 2.61 = \sqrt{\left<x^2\right>-\frac{1}{36}\left(35+x\right)^2}

    But I've no idea how to get \left<x^2\right> .
    Last edited by Captcha; November 29th 2009 at 09:47 PM.
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  4. #4
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    Quote Originally Posted by Captcha View Post
    It's not \left<x\right>^2 I want to find. It's \left<x^2\right><br />
.

    Basically I'm trying to solve for x the equation:

    2.61 = \sqrt{\left<x^2\right>-\left<x\right>^2} to find the value of x, where 2.61 is the \text{standard deviation} of the data 8,7, 4, 10, 6, x.

    I know that \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right).

     \Rightarrow \left<x^2\right> = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2.

    So 2.61 = \sqrt{\left<x^2\right>-\frac{1}{36}\left(35+x\right)^2}

    But I've no idea how to get \left<x^2\right> .
    s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }.
    \left<x^2\right>  = \frac{\displaystyle\sum_{i=1}^n x_i^2}{n}?
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  6. #6
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    Just tell me how to calculate, or give me definition of, \left<x^2\right> and I will take care of the rest.
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  7. #7
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    Quote Originally Posted by Captcha View Post
    Just tell me how to calculate, or give me definition of, \left<x^2\right> and I will take care of the rest.
    Well, I gave you a formula that will do the job of efficiently answering the question you're working on. But, if you insist:

    <X^2> = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    Well, I gave you a formula that will do the job of efficiently answering the question you're working on. But, if you insist:

    <X^2> = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}.
    Thanks a million. You are Super Fantastic!
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